Question

Seven years before, father's age was three times
of son's age .After seven years father's age will
be double that of the son's age. What is the age
of father now?​

Answer 1

Answer:

49 years

Step-by-step explanation:

Let current age of father be 'x' years

Let current age of son be 'y' years

Age of father seven years ago= x-7

Age of son seven years ago= y-7

Therefore, by the problem,

x-7= 3*(y-7)

=> x-7= 3y-21

=> x-3y= -14

=> 3y-x=14 ------(1)

Age of father after seven years= x+7

Age of son after seven years= y+7

Therefore, by the problem,

x+7= 2*(y+7)

=> x+7= 2y + 14

=> x-2y= 7 -----(2)

3y-x= 14 ---(1)

x-2y= 7 ---(2)

Equation (1) + equation (2) we get,

3y - 2y =21

=> y= 21

Putting value of y in any equation ( let's put in equation 2 here),

we get,

x- 2*21=7

=> x= 42+7

=> x=49

Current age of father = 49 years

( In questions like these on ages...always consider present ages to be x and y, that makes the problem easy to interpret and solve)

Answer 2

The age  of father now is 49 years.

Step-by-step explanation:

Given:

Seven years before, father's age was three times of son's age.

After seven years father's age will be double that of the son's age.

To Find:

The age  of father now.

Solution:

Let  Son's  present age is p  and  father's  present age is q.

As given- seven years before, father's age was three times of son's age.

    $3(p-7)=q-7$

    $3p-21=q-7$

    $3p=q+14$  -------- equation no.01.

As given-after seven years father's age will be double that of the son's age.

$2(p+7)=q+7$

$2p+14=q+7$

$2p=q-7$   --------- equation no.02.

Putting the value of p form equation no.02 to equation no.01, we get.

$3\times\frac{(q-7)}{2} =q+14$

$3(q-7)=2(q+14)$

$3q-21=2q+28$

$3q-2q=28+21$

$q=49 \ years$

Thus,the age  of father now is 49 years.

PROJECT CODE#SPJ3